Answer: [tex]2.98(10)^{10} J[/tex]
Explanation:
The work [tex]W[/tex] done by the gravitational force is given by:
[tex]W=-\Delta U=-(-\frac{GMm}{d} - \frac{GMm}{D})[/tex] (1)
Where:
[tex]\Delta U[/tex] is the variation in potential energy
[tex]G=6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}[/tex] is the Gravitational constant
[tex]M=5.972(10)^{24} kg[/tex] is the mass of the Earth
[tex]m=478 kg[/tex] is the mass of the spacecraft
[tex]d=6371(10)^{3} m[/tex] is the distance from the center of the Earth to its surface
[tex]D=7.529(10)^{12} m[/tex] is the distance between the Earth and Pluto
Rearranging (1):
[tex]W=GMm(\frac{1}{d} + \frac{1}{D})[/tex] (2)
[tex]W=(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.972(10)^{24} kg)(478 kg)(\frac{1}{6371(10)^{3} m} + \frac{1}{7.529(10)^{12} m})[/tex] (3)
Finally:
[tex]W=2.98(10)^{10} J[/tex]
